# Arc connectedness of product spaces

Is arc connected-ness well-behaved with respect to products?

That is -

$$\prod X_\alpha$$ is arc connected iff $$X_\alpha$$ is arc connected $$\forall \alpha$$

In this question on MathStackexchange, an answer is provided only for the reverse implication, that is -

If $$X_\alpha$$ is arc connected $$\forall \alpha$$, then $$\prod X_\alpha$$ is arc connected

However, I wasn't able to get an answer for the forward implication, nor have I been able to find it in any book or from Googling. So, is the forward implication true, or is there a counterexample for the same?

• They are equivalent when $X_\alpha$ is Hausdorff because in this context path-connected implies arc-connected. I will now assume that you are asking about T$_1$ spaces. – D.S. Lipham Aug 8 '20 at 19:00
• In fact, I'm asking for general $X_\alpha$, which need not even be $T_0$. – Ishan Deo Aug 8 '20 at 19:01

This is false when the spaces are not Hausdorff. Let $$X$$ be the line with two origins $$\{O_1,O_2\}$$ and $$Y$$ be the usual line. Then $$X\times Y$$ is arc connected because you can pick an arc that starts at $$(O_1,y_1)$$ travels outside $$\{O_1,O_2\}\times Y$$ and then comes back to $$(O_2,y_2)$$, but $$X$$ itself is not arc connected.
• Sorry it's early in the morning and I might be missing something obvious, but writing $X=\mathbb{R}\sqcup\mathbb{R}/\sim$ why can't I take the path $[0,\frac{1}{2}]\to[0,\frac{1}{2}]$ on the first copy of $\mathbb{R}$ and $[\frac{1}{2},1]\to[\frac{1}{2},0]$ on the second copy to link the two origins? – Daniel Robert-Nicoud Aug 9 '20 at 6:17
• @DanielRobert-Nicoud that shows that it is path connected, which is different from arc connected. You would need your map from $[0,1]\to X$ to be injective. – Gjergji Zaimi Aug 9 '20 at 6:21